Theo đề bài ta có:

\(4\le4^x\le64\)

=>\(4^1\le4^x\le4^4\)

Suy ra \(x\in\left\{1,2,3,4\right\}\)

Chúc bạn học tốt!!!

\(4^{15}\cdot9^{15}\le2^x\cdot3^x\le18^{16}\cdot2^{16}\)

\(\Rightarrow\left(2^2\right)^{15}\cdot\left(3^2\right)^{15}\le2^x\cdot3^x\le\left(2\cdot3^2\right)^{16}\cdot2^{16}\)

\(\Rightarrow2^{30}\cdot3^{30}\le2^x\cdot3^x\le2^{32}\cdot3^{32}\)

\(\Rightarrow x=31\)

tao làm nhanh

B) \(1< 3^n< 81\Rightarrow1< 3^n< 3^4\Leftrightarrow n\in\left\{1;2;3\right\}\)

C) \(4\le2^n\le64\Rightarrow2^2\le2^n\le2^6\Leftrightarrow n\in\left\{2;3;4;5;6\right\}\)

D) \(4\le4^n\le256\Rightarrow4^1\le4^n\le4^4\Leftrightarrow n\in\left\{1;2;3;4\right\}\)

phần A thì mình chịu

x_<2--> x+1/2_<5/2 mà -|x-2/3|_<0 nên Max N = 5/2 khi và chỉ khi x=2

\(-\left|x-\frac{2}{3}\right|\le0\Rightarrow\frac{1}{2}-\left|x-\frac{2}{3}\right|\le\frac{1}{2}\)

\(\Rightarrow x+\frac{1}{2}-\left|x-\frac{2}{3}\right|\le\frac{1}{2}+x\le\frac{1}{2}+2=\frac{5}{2}\)

Dấu "=" xảy ra <=> x=2/3

Vậy MaxN=5/2 <=>x=2/3

Để N đạt GTLN <=> x lớn nhất hoặc |x - 2/3| nhỏ nhất

\(\Leftrightarrow\orbr{\begin{cases}x=2\\\left|x-\frac{2}{3}\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{2}{3}\end{cases}}\)

Suy ra MaxN=\(x+\frac{1}{2}-\left|x-\frac{2}{3}\right|=2+\frac{1}{2}-\left|2-\frac{2}{3}\right|=\frac{7}{6}\)

Vậy \(MaxN=\frac{7}{6}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)

Phần dưới mình nhầm nha :v

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

a,\(8< 2^x\le2^9.2^{-5}\)

\(2^3< 2^x\le2^4\)

\(\Rightarrow x=4\)

b, \(27< 81^3.3^x< 243\)

\(3^3< 3^{12-x}< 3^5\)

\(\Rightarrow3< 12-x< 5\)

12-x=4

x=8

c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)

\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)

\(\Rightarrow x>5\)

x=6;7;8........

tìm x, biết:

(5x+1)^2=36/49